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Gen.Physics-1-2nd-Quarter-Final

Page 1: Terminal Velocity and Gravitational Force

Terminal Velocity

  • Definition: Is the constant speed a freely falling object reaches when the resistance of the medium (like air or water) prevents further acceleration.

  • Key Points:

    • At terminal velocity, the forces are balanced:

      • The force of gravity pulling the object down is equal to the force of air resistance acting upwards.

      • The object ceases to accelerate and continues falling at a constant velocity.

Gravitational Force

  • Formula:

    • ( F_{grav} = mg ) Where:

    • ( m ): mass of the object (kg)

    • ( g ): acceleration due to gravity (approx. 9.8 m/s² on Earth)

Page 2: Sample Problem 1

Problem Statement

  • Determine the velocity of a rock dropped from a bridge with an initial velocity of 15 m/s after 3.0 seconds.

Given Values

  • Initial velocity (( v_0 )): 15 m/s [down] ((=-15 m/s))

  • Acceleration (( a )): ( g = 9.8 , \text{m/s}^2 ) [down] ((=-9.8 , \text{m/s}^2))

  • Time (( , t = 3.0 , s ))

Solution Steps

  • Calculate final velocity:

    • ( v = v_0 + a imes t )

    • ( v = -15 , m/s + (-9.8 , m/s^2)(3.0 , s) )

    • ( v = -15 - 29.4 , m/s )

    • ( v = -44.4 , m/s ) or 44 m/s [down]

Page 3: Sample Problem 2

Problem Statement

  • Determine the velocity of a baseball thrown upwards with an initial velocity of 15 m/s after 2.0 seconds.

Given Values

  • Initial velocity (( v_0 )): 15 m/s [up] ((=15 m/s))

  • Acceleration (( a )): (=-9.8 , m/s^2) [down]

  • Time (( , t = 2.0 , s ))

Solution Steps

  • Calculate final velocity:

    • ( v = v_0 + a imes t )

    • ( v = 15 , m/s + (-9.8 , m/s^2)(2.0 , s) )

    • ( v = 15 - 19.6 , m/s )

    • ( v = -4.6 , m/s )

Page 4: Simple Pendulum

Definition of Simple Pendulum

  • A simple pendulum consists of an object (the bob) attached to a string that swings back and forth.

Key Characteristics

  • Amplitude: The maximum displacement from the the pendulum's rest position when swinging.

  • Period: The time needed for the pendulum to make one complete cycle of motion.

Page 5: Escape Velocity

Definition

  • Escape Velocity is the minimum speed required for an object to break free from a planet's gravitational pull without additional propulsion.

Key Concept

  • At this speed, an object possesses sufficient energy to escape the gravitational attraction and travel away indefinitely.

  • Example: The speed required for a spacecraft to leave Earth.

Page 6: Kepler's Laws

1. Law of Orbits (The Law of Ellipses)

  • Describes that planetary orbits are elliptical, not circular, with the Sun located at one focus of the ellipse.

2. Law of Areas (The Law of Equal Areas)

  • States that planets move faster when closer to the Sun (perihelion) and slower at greater distances (aphelion).

  • The area swept out over equal times remains constant.

3. Law of Periods (The Law of Harmonies)

  • Indicates that the time taken for a planet to complete an orbit increases with distance from the Sun.

  • For instance, Neptune has a far longer orbital period than Earth due to its greater distance from the Sun.

Page 7: Force of Gravity

Definition

  • The force that attracts objects with mass towards each other.

  • On Earth, it keeps us grounded and causes dropped objects to fall.

Example Calculation

  • Calculate the force of gravity for a car with a mass of 1,250 kg:

    • ( F = mg = (1,250 kg)(9.8 N/kg) = 12,250 N )

Page 8: Mass and Weight

Definitions

  • Mass: Measured in kilograms (kg); the amount of matter in an object.

  • Weight: Measured in Newtons (N); the force of gravity acting on an object.

Key Point

  • The gravitational force depends on the masses of the objects involved and the distance between them. Larger masses create stronger gravitational pulls.

Page 9: Formula for Gravitational Force

Gravitational Force Equation

  • ( F = rac{m_1 m_2}{r^2} )

    • F: Gravitational force (N)

    • m1, m2: Masses of the objects (kg)

    • r: Distance between the objects (m)

    • G: Gravitational constant ( G = 6.67 imes 10^{-11} , N , m^2/kg^2 )

Page 10: Gravitational Force Example

Problem Statement

  • Determine the gravitational force between two objects of equal mass (0.50 kg) that are 10 cm apart.

Given Values

  • Masses (( m_1 = m_2 = 0.50 kg ))

  • Distance (10 cm = 0.10 m)

  • Gravitational Constant (( G = 6.67 imes 10^{-11} , N , m^2/kg^2 ))

Solution Steps

  • ( F = G rac{m_1 m_2}{r^2} )

  • ( F = 6.67 imes 10^{-11} imes rac{(0.50)(0.50)}{(0.10)^2} = 1.7 imes 10^{-9} N )

Page 11: Homework Problems

Problem 1

  • Calculate the gravitational force between two objects with masses of 10 kg and 20 kg that are 2 meters apart.

Problem 2

  • Determine the gravitational force between Earth and Moon given:

    • Mass of Earth = ( 5.97 imes 10^{24} kg )

    • Mass of Moon = ( 7.35 imes 10^{22} kg )

    • Average distance between them = ( 3.84 imes 10^8 m )

Page 12: Hooke’s Law

Definition

  • Hooke's Law states that the force needed to stretch or compress a spring is proportional to the distance it is stretched or compressed within its elastic limit.

Formula

  • ( F = kx )

    • F: Force applied to the spring (N)

    • k: Spring constant (measure of stiffness)

    • x: Displacement from natural position (m)

Page 13: Sample Problems using Hooke’s Law

Problem 1

  • Given: Spring constant ( k = 200 , N/m ), displacement ( x = 0.15m )

  • Calculation:

    • ( F = kx = (200 , N/m)(0.15 , m) = 30N )

Problem 2

  • Given: Spring constant ( k = 48 , N/m ), mass = 0.25 kg.

  • Calculation:

    • Weight (Force) ( F = (0.25 , kg)(9.8 , N/kg) = 2.4 , N )

    • Displacement ( x = rac{F}{k} = rac{2.4 , N}{48 , N/m} = 0.050 , m , (5.0 , cm) )

Page 14: Friction

Definition

  • Friction is the force opposing the relative motion or tendency between two surfaces in contact.

Types of Friction

  1. Static Friction:

    • Resists the start of motion (e.g., a book on a table).

  2. Kinetic Friction:

    • Opposes motion between sliding surfaces (e.g., sledding down a hill).

  3. Rolling Friction:

    • Resists rolling motion (e.g., wheels on a road).

Page 15: Continued Friction Discussion

  1. Fluid Friction:

    • Resistance an object experiences when moving through a fluid (e.g., swimming).

Resultant Force

  • The overall force acting on an object when all forces are combined.

  • Determines if an object remains stationary, moves constantly, or accelerates.

Examples

  1. Determine the resultant force on a block of wood with forces of 10 N [East] and 20 N [East].

    • Answer: 30 N [East].

  2. Determine the net force on a rope if one pulls with 20 N [East] and another pulls with 5 N [West].

    • Answer: 15 N [East].

MA

Gen.Physics-1-2nd-Quarter-Final

Page 1: Terminal Velocity and Gravitational Force

Terminal Velocity

  • Definition: Is the constant speed a freely falling object reaches when the resistance of the medium (like air or water) prevents further acceleration.

  • Key Points:

    • At terminal velocity, the forces are balanced:

      • The force of gravity pulling the object down is equal to the force of air resistance acting upwards.

      • The object ceases to accelerate and continues falling at a constant velocity.

Gravitational Force

  • Formula:

    • ( F_{grav} = mg ) Where:

    • ( m ): mass of the object (kg)

    • ( g ): acceleration due to gravity (approx. 9.8 m/s² on Earth)

Page 2: Sample Problem 1

Problem Statement

  • Determine the velocity of a rock dropped from a bridge with an initial velocity of 15 m/s after 3.0 seconds.

Given Values

  • Initial velocity (( v_0 )): 15 m/s [down] ((=-15 m/s))

  • Acceleration (( a )): ( g = 9.8 , \text{m/s}^2 ) [down] ((=-9.8 , \text{m/s}^2))

  • Time (( , t = 3.0 , s ))

Solution Steps

  • Calculate final velocity:

    • ( v = v_0 + a imes t )

    • ( v = -15 , m/s + (-9.8 , m/s^2)(3.0 , s) )

    • ( v = -15 - 29.4 , m/s )

    • ( v = -44.4 , m/s ) or 44 m/s [down]

Page 3: Sample Problem 2

Problem Statement

  • Determine the velocity of a baseball thrown upwards with an initial velocity of 15 m/s after 2.0 seconds.

Given Values

  • Initial velocity (( v_0 )): 15 m/s [up] ((=15 m/s))

  • Acceleration (( a )): (=-9.8 , m/s^2) [down]

  • Time (( , t = 2.0 , s ))

Solution Steps

  • Calculate final velocity:

    • ( v = v_0 + a imes t )

    • ( v = 15 , m/s + (-9.8 , m/s^2)(2.0 , s) )

    • ( v = 15 - 19.6 , m/s )

    • ( v = -4.6 , m/s )

Page 4: Simple Pendulum

Definition of Simple Pendulum

  • A simple pendulum consists of an object (the bob) attached to a string that swings back and forth.

Key Characteristics

  • Amplitude: The maximum displacement from the the pendulum's rest position when swinging.

  • Period: The time needed for the pendulum to make one complete cycle of motion.

Page 5: Escape Velocity

Definition

  • Escape Velocity is the minimum speed required for an object to break free from a planet's gravitational pull without additional propulsion.

Key Concept

  • At this speed, an object possesses sufficient energy to escape the gravitational attraction and travel away indefinitely.

  • Example: The speed required for a spacecraft to leave Earth.

Page 6: Kepler's Laws

1. Law of Orbits (The Law of Ellipses)

  • Describes that planetary orbits are elliptical, not circular, with the Sun located at one focus of the ellipse.

2. Law of Areas (The Law of Equal Areas)

  • States that planets move faster when closer to the Sun (perihelion) and slower at greater distances (aphelion).

  • The area swept out over equal times remains constant.

3. Law of Periods (The Law of Harmonies)

  • Indicates that the time taken for a planet to complete an orbit increases with distance from the Sun.

  • For instance, Neptune has a far longer orbital period than Earth due to its greater distance from the Sun.

Page 7: Force of Gravity

Definition

  • The force that attracts objects with mass towards each other.

  • On Earth, it keeps us grounded and causes dropped objects to fall.

Example Calculation

  • Calculate the force of gravity for a car with a mass of 1,250 kg:

    • ( F = mg = (1,250 kg)(9.8 N/kg) = 12,250 N )

Page 8: Mass and Weight

Definitions

  • Mass: Measured in kilograms (kg); the amount of matter in an object.

  • Weight: Measured in Newtons (N); the force of gravity acting on an object.

Key Point

  • The gravitational force depends on the masses of the objects involved and the distance between them. Larger masses create stronger gravitational pulls.

Page 9: Formula for Gravitational Force

Gravitational Force Equation

  • ( F = rac{m_1 m_2}{r^2} )

    • F: Gravitational force (N)

    • m1, m2: Masses of the objects (kg)

    • r: Distance between the objects (m)

    • G: Gravitational constant ( G = 6.67 imes 10^{-11} , N , m^2/kg^2 )

Page 10: Gravitational Force Example

Problem Statement

  • Determine the gravitational force between two objects of equal mass (0.50 kg) that are 10 cm apart.

Given Values

  • Masses (( m_1 = m_2 = 0.50 kg ))

  • Distance (10 cm = 0.10 m)

  • Gravitational Constant (( G = 6.67 imes 10^{-11} , N , m^2/kg^2 ))

Solution Steps

  • ( F = G rac{m_1 m_2}{r^2} )

  • ( F = 6.67 imes 10^{-11} imes rac{(0.50)(0.50)}{(0.10)^2} = 1.7 imes 10^{-9} N )

Page 11: Homework Problems

Problem 1

  • Calculate the gravitational force between two objects with masses of 10 kg and 20 kg that are 2 meters apart.

Problem 2

  • Determine the gravitational force between Earth and Moon given:

    • Mass of Earth = ( 5.97 imes 10^{24} kg )

    • Mass of Moon = ( 7.35 imes 10^{22} kg )

    • Average distance between them = ( 3.84 imes 10^8 m )

Page 12: Hooke’s Law

Definition

  • Hooke's Law states that the force needed to stretch or compress a spring is proportional to the distance it is stretched or compressed within its elastic limit.

Formula

  • ( F = kx )

    • F: Force applied to the spring (N)

    • k: Spring constant (measure of stiffness)

    • x: Displacement from natural position (m)

Page 13: Sample Problems using Hooke’s Law

Problem 1

  • Given: Spring constant ( k = 200 , N/m ), displacement ( x = 0.15m )

  • Calculation:

    • ( F = kx = (200 , N/m)(0.15 , m) = 30N )

Problem 2

  • Given: Spring constant ( k = 48 , N/m ), mass = 0.25 kg.

  • Calculation:

    • Weight (Force) ( F = (0.25 , kg)(9.8 , N/kg) = 2.4 , N )

    • Displacement ( x = rac{F}{k} = rac{2.4 , N}{48 , N/m} = 0.050 , m , (5.0 , cm) )

Page 14: Friction

Definition

  • Friction is the force opposing the relative motion or tendency between two surfaces in contact.

Types of Friction

  1. Static Friction:

    • Resists the start of motion (e.g., a book on a table).

  2. Kinetic Friction:

    • Opposes motion between sliding surfaces (e.g., sledding down a hill).

  3. Rolling Friction:

    • Resists rolling motion (e.g., wheels on a road).

Page 15: Continued Friction Discussion

  1. Fluid Friction:

    • Resistance an object experiences when moving through a fluid (e.g., swimming).

Resultant Force

  • The overall force acting on an object when all forces are combined.

  • Determines if an object remains stationary, moves constantly, or accelerates.

Examples

  1. Determine the resultant force on a block of wood with forces of 10 N [East] and 20 N [East].

    • Answer: 30 N [East].

  2. Determine the net force on a rope if one pulls with 20 N [East] and another pulls with 5 N [West].

    • Answer: 15 N [East].

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